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The weight of oranges growing in an orchard is normally distributed with a mean weight of 5 oz. And a standard deviation of 1 oz. Using the empirical rule, determine what interval would represent weights of the middle 95% of all oranges from this orchard.

Respuesta :

Answer:

The interval would represents weights of the middle 95% of all oranges from this orchard.

95%of confidence interval is determined by    μ ± 2σ

(5-2 , 5+2)

(3 ,7)

Step-by-step explanation:

Empirical rule:-

The empirical rule states that for a normal distribution

  • 68% of data falls within the first standard deviation from the mean
  • 95% fall within two standard deviations
  • 99.7% fall within three standard deviations

This empirical rule is called the 68-95-99.7 rule

68% of confidence interval is determined by

                                                                     μ ± σ

95%of confidence interval is determined by

                                                                     μ ± 2σ

99.7%of confidence interval is determined by

                                                                     μ ± 3σ

Given data

Mean of the Population  μ  = 5 o z

Standard deviation of the Population  σ = 1 o z

95%of confidence interval is determined by

                                                                     μ ± 2σ

                                                                     5 ± 2(1)

The interval would represents weights of the middle 95% of all oranges from this orchard.

(5 -2 , 5+2)

(3 , 7)

cf1116rcsa

Answer:

(2.5,8.5)

Step-by-step explanation:

Have a good day :)

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